Question: Which integral gives the arc length of the curve $y=e^{3x}$ over the interval $[0, 2]$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{0}^{2} \sqrt{1+6e^{6x}}\,dx$ (Choice B) B $ \int_{0}^{2} \sqrt{1+6e^{9x}}\,dx$ (Choice C) C $ \int_{0}^{2} \sqrt{1+9e^{9x}}\,dx$ (Choice D) D $ \int_{0}^{2} \sqrt{1+9e^{6x}}\,dx$
Answer: The arc length $L$ of the curve $y=f(x)$ over the interval $[a, b]$ is $ L=\int_a^b\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\, dx$. First, calculate $dy/dx$. $\begin{aligned} y &= e^{3x}\\ \\ \dfrac{dy}{dx} &= 3e^{3x} \end{aligned}$ Now apply the arc length formula on the interval $[0, 2]$ and simplify the integral. $\begin{aligned} L &= \int_{0}^{2} \sqrt{1+(3e^{3x})^2}\,dx \\\\ &= \int_{0}^{2} \sqrt{1+9e^{6x}}\,dx \end{aligned}$